The Stacks project

Proof. This proof is the same as the proof of Lemma 93.12.4 but easier. We suggest the reader to skip the proof. The map is given by localization: given $0 \to N \to C \to B \to 0$ in $\text{Exal}_ k(B, N)$ we send it to the localization $S_ C^{-1}C$ of $C$ with respect to the inverse image $S_ C \subset C$ of $S$. Compare with the proof of Lemma 93.8.7.

The smooth locus of a morphism of schemes is open by definition. Let $J \subset B$ be an ideal cutting out the set of points in $\mathop{\mathrm{Spec}}(B)$ where $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is not smooth. Since $k \to B$ is of finite presentation the complex $\mathop{N\! L}\nolimits _{B/k}$ can be represented by a complex $N^{-1} \to N^0$ where $N^ i$ is a finite $B$-module, see Algebra, Section 10.134 and in particular Algebra, Lemma 10.134.2. As $B$ is Noetherian, this means that $\mathop{N\! L}\nolimits _{B/k}$ is pseudo-coherent. For $g \in J$ the $k$-algebra $B_ g$ is smooth and hence $(\mathop{N\! L}\nolimits _{B/k})_ g = \mathop{N\! L}\nolimits _{B_ g/k}$ is quasi-isomorphic to a finite projective $B$-module sitting in degree $0$. Thus $\text{Ext}^ i_ B(\mathop{N\! L}\nolimits _{B/k}, N)_ g = 0$ for $i \geq 1$ and any $B$-module $N$. Finally, we have

The first equality by More on Algebra, Lemma 15.99.2 and Algebra, Lemma 10.134.13. The second because $\text{Ext}^1_ B(\mathop{N\! L}\nolimits _{B/k}, N)$ is $J$-power torsion and elements of $S$ act invertibly on $J$-power torsion modules. This concludes the proof by the description of $\text{Exal}_ A(B, N)$ as $\text{Ext}^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ given just above Lemma 93.12.4. $\square$